链表的部分翻转

给定一个链表，翻转该链表从m到n的位置。要求直接翻转而非申请新空间。

#include <iostream>#include <vector>#include <algorithm>#include <map>#include <cstdio>using namespace std;typedef struct tagSNode{int  value;tagSNode* pNext;tagSNode(int v) :value(v), pNext(NULL){}}SNode;void Reverse(SNode* pHead, int from, int to){SNode* pCur = pHead->pNext;int i;for (i = 0; i<from - 1; i++){pHead = pCur;pCur = pCur->pNext;}SNode* pPre = pCur;pCur = pCur->pNext;to--;SNode* pNext;for (; i<to; i++){//此处需要结合图示pNext = pCur->pNext;pCur->pNext = pHead->pNext;pHead->pNext = pCur;pPre->pNext = pNext;pCur = pNext;}}void Destroy(SNode* p){SNode* next;while (p){next = p->pNext;delete p;p = next;}}int main(){SNode* pHead = new SNode(0);int i;for (i = 0; i<10; i++){SNode* p = new SNode(rand() % 100);p->pNext = pHead->pNext;pHead->pNext = p;}//Print(pHead);Reverse(pHead, 4, 8);//Print(pHead);//Print(pHead);Destroy(pHead);return 0;}